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Proof step 2 from proof gcd-less (previous step)

i1 | i, i1 | j, gcd(i, j) < i1 ⊦ 0 = i

User applied rule insert lemma

And obtained the following new goals

Proof step 3:
⊦ i0 | i ∧ i0 | j ∧ (i ≠ 0 ∨ j ≠ 0) → i0 ≤ gcd(i, j)

Rule argument

spec: int-gcd inst: name: gcd-max seq: ⊦ i0 | i ∧ i0 | j ∧ (i ≠ 0 ∨ j ≠ 0) → i0 ≤ gcd(i, j) subst: [i0, i, j] → [i1, i, j] flags: current, []

Additional information

Used simplifier rules: lels-03: ⊦ i < j → (¬ ¬ j ≤ i ↔ false); from specification int-abs Used terms: i1, i, j